# What is the oxidation state of the potassium permanganate?

Feb 16, 2015

I assume the question refers to the oxidation number of manganese, or $M n$, in potassium permanganate, or $K M n {O}_{4}$.

If you're familiar with the rules for calculating oxidation numbers (more here: http://socratic.org/questions/how-do-you-calculate-oxidation-numbers), you'll notice immediately that each oxygen atom has an O.N. of -2 and that the potassium atom has an O.N. of +1.

Since $K M n {O}_{4}$ is a neutral compound, the sum of the O.N. calculated for each atom must be equal to zero. This implies that

$O {N}_{K} + O {N}_{M n} + 4 O {N}_{O} = 0$, or

$\left(+ 1\right) + O {N}_{M n} + 4 \cdot \left(- 2\right) = 0 \implies O {N}_{M n} = 0 - 1 + 8$

$O {N}_{M n} = + 7$

As a side note, potassium permanganate is often used in experiments that show the color changes that accompany the reduction of manganese from +7 to +6, and from +6 to +4.

This is done by reacting $K M n {O}_{4}$ with a solution containing sodium hydroxide and sucrose (regular table sugar). The rezulting solution will go from being purple (ON +7), to being green (ON +6) and finally to being yellow (ON +4).

Here's a video on such an experiment: